00 Assignment Presentasi 2014 Edisi 4 Senini 18-8-2014

INTERACTION THEORY

[NEW PARADIGM] FOR SOLVING

THE ASSIGNMENT PROBLEM

Department of Industrial Engineering Institut Teknologi Bandung,Indonesia

[email protected]

Anang Zaini Gani

Bandung, August 20-21th, 2014

Department of Industrial Engineering Widyatama University,Indonesia [email protected]

INTERACTION THEORY

[NEW PARADIGM] FOR

SOLVING THE

ASSIGNMENT PROBLEM

Anang Zaini Gani

Department of Industrial Engineering

Institut Teknologi Bandung,Indonesia

[email protected]

Department of Industrial Engineering

Widyatama University,Indonesia

[email protected]

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INTRODUCTION

OBJECTIVETHE ASSIGNMENT

PROBLEM

INTERACTION THEORY

COMPUTATIONAL EXAMPLES

NEW PARADIGM

CONCLUSION

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I. INTRODUCTION

The Assignment Problem is a problem of relationship

or interaction between two independent groups. The

first group is called worker or assignee (A) and the

second group is task or job (J). . The purpose is to find

a combination, that gives an optimal solution

(minimum or maximum).

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Objective : how to assign the jobs to the workers to minimize the cost?

Constraint : each worker need perform only one job and each job need be assigned to only

one worker.

n! possible assignment sets (10! = 3,628,800), (100! = 9.3 10157)

Job 1 Job 2 Job 3 Job 4

Worker A 1 4 8 12 16

Worker A2 22 14 6 8

Worker A3 13 3 9 12

Worker A4 7 8 12 5

Workers: A1, A2, A3, A4 Jobs: 1, 2, 3, 4.

Cost matrix:

Total Cost : (A1-j1 = 4) + (A2-j3 = 6) + (A3-j2 = 3) + (A4-j4 = 5) = 18

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II. OBJECTIVE

A new method for solving assignment

problem with a simple yet effective and efficient to solve all kinds of assignment problems (unbalanced assignment problem , with incomplete data and minimize and maximize objective)

III. THE ASSIGNMENT

PROBLEM

The assignment problem is a simple Combinatorial

Optimization Problem

A. Formal

Mathematical

Definition

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1. BRUTE-FORCE METHOD

a) Enumerate all candidates sets

b) n! possible assignment sets

c) Exponential runtime complexity

2. SIMPLEX METHOD

a) More variables

b) May lead to more iterations.

c) Inefficient.

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3. HUNGARIAN METHOD

Hungarian Method is one of the methods that get mostly used.

Hungarian Method was developed by Harold Kuhn in 1955. The time complexity of the original algorithm is

O(n4). Edmonds and Karp, and Tomizawa modified to

achieve an O(n3) running time.

There are restrictions to Hungarian Method, then the complicated iteration process occurs or limited in its

usability. (square matrix, dummy, incomplete data)

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4. OTHER METHODS

There are several other algorithms used to solve the

assignment problem. be very complicated.

5. INTERACTION THEORY

The algorithm is simple yet effective and efficient to

solve all kinds of assignment problems such as

unbalanced assignment problem, and minimize and

maximize objective. Moreover, the algorithm

sometimes produces more than one optimal

solution.

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IV. INTERACTION THEORY

In 1965 Anang Z. Gani [28] did research on the Facilities Planning

and Traveling Salesman Problem as a special project

(Georgia Tach in 1965).Supervision James Apple

Later, J. M. Devis and K. M. Klein further continued the original

work of Anang Z. Gani

Then M. P. Deisenroth PLANET direction of James Apple (Georgia Tech in1971)

Since 1966, Anang Z. Gani has been continuing his research and

further developed a new concept which is called The Interaction Theory (INSTITUT TEKNOLOGI BANDUNG)

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1) Basic Concept

The Interaction Theory resulted in

development of the model (matrix)

representing the flow in general

The absolute value or the number of a

element as an individual of a matrix can not

be used in priority setting

the Asignmemt matrix has two values,

1. the initial absolute value (interaction

value)

2. the relative value (interaction coefficient)

DIM = The Delta Interaction Matrix

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1 2 3 4

1 0 700 10 20

2 2 0 800 15

3 4 3 0 10

4 10 2 30 0

RELATIVE VALUE

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14

Two parallel lines

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Two parallel lines distorted

(Hering illusion)

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The formula for the interaction coefficient

( ci,j ) is:

ci,j = xi,j2/(Xi. .X.j).

Xi. =

m

j 1

xij (i = 1 . m )

X.j =

n

i 1

xij (j = 1 . n )

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The Interaction Theory has wide range of applications

from solving hard problems Traveling Salesman

Problem (TSP), Vehicle Routing Problem (VRP),

Clustering, Transportation Problem to simple problem

such as The Assignment Problem.

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TABLE 2. THE INTERACTION MATRIX (IMAT)

To

From

I

J

N

Total

1 x11 x1j xin Xi.

I xii x1j xin Xi.

M xm1 Xmn

Total X.j Xi..

2) The Proposed Method (Algorithm)

1) The interaction matrix (IMAT)

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TABLE 3.THE INTERACTION COEFFICIENT MATRIX (ICOM)

To

From

i

j

n

1 c11 Cij cin

i ci1 cij cin

m cm1 cmj cmn

2) Normalization of IMAT (NIMAT)

The formula for the interaction coefficient is:

ci,j= xi,j2/(Xi. .X.j)

3) The Interaction Coefficient Matrix (ICOM)

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4) The Smallest Element

5) Unique PSE (Potential Solution Element)

6) Comparison between 2 pairs

If PSE(1)+PSE(3) is less than PSE(2)+PSE(4), then PSE(1) is defined as a solution.

If PSE(1)+PSE(3) is more than PSE(2)+PSE(4), then PSE(2) is defined as a solution.

The next process is to compile a new ICOM

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three applications of the algorithm :

1. standard problem with more than one

solution

2. unequal matrix

3. incomplete data problem

V. COMPUTATIONAL EXAMPLES

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1 2 3 4

Ship

1 500 400 600 700

2 600 600 700 500

3 700 500 700 600

4 500 400 600 600

TABLE 1. THE INTERACTION MATRIX (IMAT)

1. Assignment Problem 4x4, From Hilliers book, 7th Edition, page 399

Four cargo ships will be used for shipping goods from one port to four other ports (labeled 1, 2, 3, 4). Any ship can be used for making any one of these four trips.

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1 2 3 4

1 10500 10400 10600 10700 42200

2 10600 10600 10700 10500 42400

3 10700 10500 10700 10600 42500

4 10500 10400 10600 10600 42100

42300 41900 42600 42400

TABLE 2. NORMALIZATION OF IMAT (NIMAT)

each elements are added by 10.000

C11= 105002 /42300 x 42200 = 617.63

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1 2 3

4

1 617.63 611.70 625.01 639.87

2 626.48 632.46 633.86 613.26

3 636.85 619.12 632.37 623.53

4 619.09 613.16 626.50 629.45

TABLE 3. THE INTERACTION COEFFICIENT MATIX (ICOM)

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4) The Smallest Element.

1 2 3 4

1 5.93 0.00 13.31 28.16

2 14.78 20.76 22.16 1.56

3 25.15 7.42 20.66 11.83

4 7.39 1.46 14.80 17.75

TABLE 4. EACH ELEMENTS SUBSTRACTED BY SMALLEST

ELEMENTS (PSE=Potential Solution Element)

PSE 1= PSE 2=

PSE 3= PSE 4=

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5) Unique PSE

Element 2-4 is a unique PSE, then 2-4 is the

solution and row 2 and column 4 is blocked

The three alternative solutions are:

1. (1-1=500)+(2-4= 500)+(3-3 =700)+(4-2= 400)=2100

2. (1-2=400)+( 2-4=500)+( 3-3=700)+(4-1=500)= 2100

3. (1-1=500)+( 2-4=500)+( 3-2=500)+( 4-3=600)=2100

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PSE 1(1-2) + PSE 3(4-1) = 7.93 with PSE 2(1-1) + PSE 4 (4-2) = 7.93.


Stroke

Carl

Chris

David

Tony

Ken

Backstroke 37.7 32.9 33.8 37 35.4

Breaststroke 43.4 33.1 42.2 34.7 41.8

Butterfly 33.3 28.5 38.9 30.4 33.6

Freestyle 29.2 26.4 29.6 28.5 31.1

TABLE 5. THE INTERACTION MATRIX (IMAT) 4x5

Note : dummy is not necessary though unbalanced matrix

2. Assignment Problem 4x5 (Hilier & Lieberman, page 399 problem 8.3-4 )

The coach of an age group swim team needs to assign swimmers to a 200-yard medley relay team to be sent to the Junior Olympics.

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1 2 3 4 5

1 10037.7 10032.9 10033.8 10037 10035.4 40141.4

2 10043.4 10033.1 10042.2 10034.7 10041.8 40153.4

3 10033.3 10028.5 10038.9 10030.4 10033.6 40131.1

4 10029.2 10026.4 10029.6 10028.5 10031.1 40113.7

40143.6 40120.9 40144.5 40130.6 40141.9

TABLE 6. NORMALIZATION OF IMAT


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1 2 3 4 5

1 625.26 625.01 624.76 625.37 625.00

2 625.78 624.85 625.62 624.90 625.61

3 624.87 624.63 625.55 624.71 624.93

4 624.63 624.64 624.67 624.75 624.89

TABLE 7. THE INTERACTION COEFFICIENT MATIX (ICOM)

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4) The Smallest Element

1 2 3 4 5

1 0.63 0.39 0.13 0.75 0.37

2 1.16 0.23 0.99 0.27 0.98

3 0.24 0.00 0.93 0.09 0.31

4 0.00 0.01 0.04 0.12 0.27

There are 6 PSE, marked by bold faces.

TABLE 8 : EACH ELEMENTS SUBSTRACTED BY SMALLEST ELEMENTS

PSE 1=

PSE 2= PSE 3=

PSE 4=

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5) Unique PSE

Element 1-3 is a unique PSE, then1-3 is

the solution and row 1 and column 3 is

blocked.


the solution:

(1-3=33.8)+(2-4=34.7)+(3-2=28.5)+(4-1=29.2)= 126.2.

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1) Original Matrix

A B C D E

1 11 19 M 19 20

2 38 M 15 14 10

3 23 17 21 M M

4 31 37 28 35 46

5 27 43 23 18 M

6 15 21 33 17 28

TABEL 9 . ORIGINAL MATRIX

3. Assignment Problem 6x5 with M (Yih-

long Chang, page 171 exercises 2)

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2). The interaction matrix (IMAT)

A B C D E

1 11 19 33 19 20

2 38 43 15 14 10

3 23 17 21 35 46

4 31 37 28 35 46

5 27 43 23 18 46

6 15 21 33 17 28

(M value taken from biggest value from column or row)

TABEL 10 . THE INTERACTION MATRIX (IMAT) 6x5

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3). Normalization of Imat

A B C D E

1 10011 10019 10033 10019 10020

2 10038 10043 10015 10014 10010

3 10023 10017 10021 10035 10046

4 10031 10037 10028 10035 10046

5 10027 10043 10023 10018 10046

6 10015 10021 10033 10017 10028

TABEL 11 . NORMALIZATION OF IMAT


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4). The Interaction Coefficient Matix (ICOM)

A B C D E

1 332.58 332.92 334.00 333.15 332.90

2 334.26 334.40 332.69 332.70 332.12

3 333.12 332.52 332.94 333.95 334.36

4 333.41 333.62 333.17 333.72 334.13

5 333.28 334.15 332.97 332.72 334.26

6 332.77 332.97 333.92 332.94 333.35

TABEL 12 . THE INTERACTION COEFFICIENT MATIX (ICOM)

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5). Each Elements Substracted By Smallest Elements

A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13 . EACH ELEMENTS SUBSTRACTED BY

SMALLEST ELEMENTS

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Element 3-B is a unique PSE, then 3-B is the solution and

row 3 and column B is blocked.

6). Unique PSE

A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13b . UNIQUE PSE (3-B)

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7). Comparison between 2 pairs

A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13c . COMPARISON BETWEEN 2 PAIRS [1]

PSE1= PSE2=

PSE3= PSE4=

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A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13d . COMPARISON BETWEEN 2 PAIRS [2]

PSE1=

PSE2= PSE3=

PSE4=

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A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13e . COMPARISON BETWEEN 2 PAIRS [3]

PSE1=

PSE2= PSE3=

PSE4=

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A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13f . COMPARISON BETWEEN 2 PAIRS [4]

PSE1=

PSE2= PSE3=

PSE4=

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A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13g . COMPARISON BETWEEN 2 PAIRS [5]

PSE1=

PSE2= PSE3=

PSE4=

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A B C D E

1 0.47 0.81 1.89 1.04 0.78

2 2.14 2.28 0.57 0.59 0.00

3 1.00 0.41 0.82 1.84 2.25

4 1.30 1.50 1.05 1.60 2.01

5 1.17 2.04 0.86 0.61 2.15

6 0.65 0.86 1.81 0.83 1.24

TABEL 13h . COMPARISON BETWEEN 2 PAIRS [6]

PSE1=

PSE2= PSE3=

PSE4=

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the total assignment cost of: (1-A=11) + (3-B=17) +

(5-C=23) + (6-D=17) + (2-E=10) equals 78.

TABLE 14. OLD PARADIGM VS NEW PARADIGM

No OLD PARADIGM NEW PARADIGM

1. Absolute value Interaction coefficient

2. Balance matrix Balance and unbalance

matrix

3. Need dummy No dummy

4. Need big M for

incomplete data

Incomplete data filled

using existing data

5. Often more than one

iteration

Only one iteration

6. Limited applications Wide applications

VI. NEW PARADIGM

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VII. CONCLUSION

THE INTERACTION THEORY HAS WIDE RANGE OF APPLICATIONS

from solving hard problems such as Traveling Salesman Problem to simple problems

such as The Assignment Problem.

INTERACTION THEORY PROVIDES A NEW PARADIGM for solving The

Assignment Problem. The algorithm is more simple yet very effective and efficient

compare to the Hungarian Method.

IT CAN BE USED TO SOLVE MXN matrix assignment problem easily and

also the problem of INCOMPLETE DATA (BIG M). It also performs well in large size

problem, and often provides MORE THAN ONE SOLUTION (minimum and maximum

solution).

This NEW PARADIGM has opened the door to WIDER APPLICATIONS and

users due to its SIMPLICITY. Overall, The Interaction Theory shows A NEW

CONCEPT which has the development in mathematics, computer science and

operations research and their applications.

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THANK YOU

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Documents

00 Assignment Presentasi 2014 Edisi 4 Senini 18-8-2014